uchihaitachi wrote:
gmatt1476 wrote:
The difference 942 − 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how many 3-digit numbers abc are possible such that the difference abc − cba is a positive multiple of 7 ?
A. 142
B. 71
C. 99
D. 20
E. 18
PS01661.01
The numbers are 100a + 10b +c and 100c + 10b + a.
After taking the difference, we are left with 99(a-c).
Now for this number to be a multiple of 7,
a-c = 7 or 0. for
(a-c) = 7, there are two possibilities ---> (9,2) and (8,1)
and for
a-c = 0, a = c, so there are 9 possibilities that are 1 to 9.
The value of b does not matter and since b cannot the value 0, so there 9 possibilities for b in each case.
There are a total of 11 cases so the total number of numbers possible are 9*11 = 99.
IMO, answer is CBy this method, the number 252, 343, 515 etc. all qualify to represent 'abc'.
Let's take 252 as abc.
cba = 252.
abc - cba = 0.
But the question wants the difference to be a
positive multiple of 7.
Now,
0 is neither positive nor negative.Thus, when abc = 252, then abc – cba is not a positive multiple of 7.
The condition
a = c is not valid.Only (a,c) = (9,2) and (8,1) is possible.
In both cases b can take 9 possible values as b ≠ 0.
2 x 9 = 18 possible choices.